Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,
U
B
=0
When the car is at point A, the potential energy of the car-Earth system is given by
U
A
=mgy
where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:
y=(41.1m)sin40.0
0
=26.4m
Thus,
U
A
=(1000kg)(9.80m/s
2
)(26.4m)=2.59∗10
5
J
The change in potential energy of the car-Earth system as the car moves from A to B is
U
B
−U
A
=0−2.59∗10
5
J=−2.59∗10
5
J
(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U
A
=0. The potential energy of the system when the car is at point B is given by U
B
=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.
Thus,
Answer: 71.72 days
Explanation:
This problem can be solved using the <u>Radioactive Half Life Formula: </u>
(1)
Where:
is the final amount of Iodine-131
is the initial amount of Iodine-131
is the time elapsed
is the half life of Iodine-131
Knowing this, let's substitute the values and find from (1):
(2)
(3)
Applying natural logarithm in both sides:
(4)
(5)
Finding :
The weight of an object is the
force with which it is attracted to earth. The gravity of an object or body of
an object is high on earth than at the atmosphere. It has an average of
gravitational constant equal to 9.8066 or 9.8 meters per second. In truth, the acceleration
of the object depend upon its location, the latitude and altitude, on
earth.
<span> </span>
Sorry I can’t give the answer cause I don’t have the graph
I’m so sorry but if u give the graph I will be able to give the answer
Answer:
hi you can just buy new ruler lol joke
1. energy and air
2.
